What is the time complexity of Watering Plants?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Watering Plants?

This problem uses Array, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Watering Plants?

Insight 1: We need to keep track of the current water level in the watering can. Insight 2: We need to determine when a refill is necessary and calculate the steps involved in going back to the river and returning to the plant.

Which companies ask LeetCode 2079. Watering Plants in interviews?

This problem is frequently asked at Amadeus.

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Watering Plants - LeetCode 2079 Solution

Watering Plants - Complete Solution Guide

Watering Plants is LeetCode problem 2079, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You want to water n plants in your garden with a watering can. The plants are arranged in a row and are labeled from 0 to n - 1 from left to right where the i th plant is located at x = i . There is a river at x = -1 that you can refill your watering can at. Each plant needs a specific amount of water. You will water the plants in the following way: Water the plants in order from left to right. After watering the current plant, if you do not have enough water to completely water the next plant,

Detailed Explanation

The problem asks us to simulate watering plants in a row using a watering can. We start at a river (x = -1) and must water plants from left to right (x = 0 to n-1). Each plant needs a specific amount of water. If we don't have enough water in our watering can to water the next plant completely, we must return to the river to refill. The goal is to calculate the total number of steps needed to water all plants, where each step represents moving one unit along the x-axis.

Solution Approach

The provided solution uses a straightforward simulation approach. It iterates through the plants array, keeping track of the current water level in the watering can. For each plant, it checks if there is enough water. If not, it adds the steps required to return to the river and refill, then adds the steps to return to the current plant. If there is enough water, it waters the plant and updates the water level. The steps are accumulated as the loop progresses.

Step-by-Step Algorithm

Step 1: Initialize `steps` to 0 and `current_water` to `capacity`.
Step 2: Iterate through the `plants` array using a `for` loop, with `i` as the index and `water_needed` as the water required for the current plant.
Step 3: Check if `current_water` is less than `water_needed`. If it is, the watering can needs to be refilled.
Step 4: If a refill is needed, add `i` to `steps` (steps to return to river), reset `current_water` to `capacity`, and add `i + 1` to `steps` (steps to reach the plant from the river).
Step 5: If a refill is not needed, simply add 1 to `steps` (one step to the next plant).
Step 6: Subtract `water_needed` from `current_water` to simulate watering the plant.
Step 7: After the loop completes, return the accumulated `steps`.

Key Insights

Insight 1: We need to keep track of the current water level in the watering can.
Insight 2: We need to determine when a refill is necessary and calculate the steps involved in going back to the river and returning to the plant.
Insight 3: The distance to the river from plant 'i' is 'i+1', and the distance back from plant 'i' to the river is 'i'.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: Amadeus.

Watering Plants | LeetCode 2079

Medium

Array Simulation

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int wateringPlants(int[] plants, int capacity) {
        int steps = 0;
        int current_water = capacity;

        for (int i = 0; i < plants.length; i++) {
            int water_needed = plants[i];
            if (current_water < water_needed) {
                // Steps to go back to the river from plant i-1
                steps += i;

                // Refill the watering can
                current_water = capacity;

                // Steps to walk from the river to the current plant i
                steps += (i + 1);
            } else {
                // Walk one step from the previous position (plant i-1 or river)
                steps += 1;
            }

            // Water the current plant
            current_water -= water_needed;
        }

        return steps;
    }
}

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Problem Info

Description

Problem 2079: Watering Plants - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

Amadeus

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2079

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `steps` to 0 and `current_water` to `capacity`.
2
Step 2: Iterate through the `plants` array using a `for` loop, with `i` as the index and `water_needed` as the water required for the current plant.
3
Step 3: Check if `current_water` is less than `water_needed`. If it is, the watering can needs to be refilled.
4
Step 4: If a refill is needed, add `i` to `steps` (steps to return to river), reset `current_water` to `capacity`, and add `i + 1` to `steps` (steps to reach the plant from the river).
5
Step 5: If a refill is not needed, simply add 1 to `steps` (one step to the next plant).
6
Step 6: Subtract `water_needed` from `current_water` to simulate watering the plant.
7
Step 7: After the loop completes, return the accumulated `steps`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm processes each plant once. The space complexity is constant because the algorithm uses a fixed number of variables regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

No alternative approach with better complexity.

Due to the need to simulate each watering step, it's unlikely to find an algorithm with time complexity better than O(n). Any alternative that attempts to pre-calculate refilling points would likely still require a linear scan of the input.

Time: N/ASpace: N/A

Trade-offs: N/A

✨

Key Insights

1Insight 1: We need to keep track of the current water level in the watering can.
2Insight 2: We need to determine when a refill is necessary and calculate the steps involved in going back to the river and returning to the plant.
3Insight 3: The distance to the river from plant 'i' is 'i+1', and the distance back from plant 'i' to the river is 'i'.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: When the capacity is smaller than the smallest plant's water requirement. The algorithm correctly handles this by refilling the can before watering each plant.
Edge case 2: When the capacity is large enough to water all plants without refilling. The algorithm still correctly calculates the total steps.
Edge case 3: Empty plants array (n=0). Although constraints state n>=1, a robust solution should handle this gracefully by returning 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to add the steps to return to the river and the steps to go back to the plant when a refill is needed. A common error is to only add the steps to return to the river.

✗Common mistake 2: Incorrectly calculating the distance back to the river or to the current plant after refilling. It's crucial to remember that the river is at x = -1, so the distance to the river from plant i is i, and the distance from the river to plant i is i+1.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: A robot watering plants in a greenhouse, where it has a limited water tank and needs to return to a refilling station periodically.

🏦Real-world scenario 2: A delivery truck making multiple stops, where the truck has limited fuel and needs to return to a depot for refueling when the fuel is insufficient for the next delivery.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Watering Plants"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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